The vapour pressure of pure liquid `M & N` are `700mm` of `Hg` and `450mm` of `Hg` respectively. Which of the following option is correct ?
Given : `X_(N),X_(M)=` mole fraction of `N & M` in liquid phase
`Y_(N),Y_(M)=` mole fraction of `N & M` in vapour phase
The vapour pressure of pure liquid `M & N` are `700mm` of `Hg` and `450mm` of `Hg` respectively. Which of the following option is correct ?
Given : `X_(N),X_(M)=` mole fraction of `N & M` in liquid phase
`Y_(N),Y_(M)=` mole fraction of `N & M` in vapour phase
Given : `X_(N),X_(M)=` mole fraction of `N & M` in liquid phase
`Y_(N),Y_(M)=` mole fraction of `N & M` in vapour phase
A
`X_(M)-X_(N)gtY_(M)-Y_(N)`
B
`(X_(M))/(X_(N))gt(Y_(M))/(Y_(N))`
C
`(X_(M))/(X_(N))lt(Y_(M))/(Y_(N))`
D
`(X_(M))/(X_(N))-(Y_(M))/(Y_(N))`
Text Solution
AI Generated Solution
The correct Answer is:
To solve the problem, we need to analyze the given vapor pressures of the pure liquids M and N, and how they relate to their mole fractions in both the liquid and vapor phases.
### Step 1: Identify the given data
- Vapor pressure of pure liquid M (P₀M) = 700 mmHg
- Vapor pressure of pure liquid N (P₀N) = 450 mmHg
### Step 2: Determine the volatility of the liquids
Since the vapor pressure of liquid M is greater than that of liquid N (P₀M > P₀N), we can conclude that liquid M is more volatile than liquid N. This means that at equilibrium, more molecules of M will escape into the vapor phase compared to N.
### Step 3: Define the mole fractions
Let:
- Xₘ = mole fraction of M in the liquid phase
- Xₙ = mole fraction of N in the liquid phase
- Yₘ = mole fraction of M in the vapor phase
- Yₙ = mole fraction of N in the vapor phase
### Step 4: Apply Raoult's Law
According to Raoult's Law, the partial vapor pressure of each component in a mixture is given by:
- Pₘ = Xₘ * P₀M
- Pₙ = Xₙ * P₀N
The total vapor pressure (P_total) of the mixture can be expressed as:
\[ P_{total} = Pₘ + Pₙ = Xₘ * P₀M + Xₙ * P₀N \]
### Step 5: Analyze the mole fractions in the vapor phase
Since M is more volatile, we expect that:
- Yₘ > Yₙ
This indicates that the mole fraction of the more volatile component (M) in the vapor phase is greater than that of the less volatile component (N).
### Step 6: Compare the ratios of mole fractions
From the above observations, we can derive the relationship between the mole fractions in the liquid and vapor phases:
- Since Yₘ > Yₙ and M is more volatile, we can infer that:
\[ \frac{Xₘ}{Xₙ} < \frac{Yₘ}{Yₙ} \]
### Conclusion
Thus, the correct option from the given choices is:
\[ \frac{Xₘ}{Xₙ} < \frac{Yₘ}{Yₙ} \]
Topper's Solved these Questions
Similar Questions
Explore conceptually related problems
Liquid 'M' and liquid 'N' form an ideal solution. The vapour pressures of pure liquids 'M' and 'N' are 450 and 700 mmHg, respectively, at the same temperature. Then correct statement is : (x_(M) = Mole fraction of 'M' in solution, x_(N) = Mole fraction of 'N' in solution , y_(M) = Mole fraction of 'M' in vapour phase , y_(N) = Mole fraction of 'N' in vapour phase)
Liquid ‘M’ and liquid ‘N’ form an ideal solution. The vapour pressures of pure liquids ‘M’ and ‘N’ are 450 and 700 mmHg, respectively, at the same temperature. Then correct statement is: ( x_(M)= Mole fraction of ‘M’ in solutions , " " x_(N)= Mole fraction of ‘N’ in solution , y_(M)= Mole fraction of ‘M’ in vapour phase , " " y_(N)= Mole fraction of ‘n’ in vapour phase)
For a ideal liquid solution with P_A^(@)gtP_B^(@) , which relation between X_(A) ((mole fraction of A in liquid phase) and Y_(A) (mole fraction of A in vapour phase) is correct ?
Vapour pressure of pure liquids P and Q are 700 and 450 mm Hg respectively at 330K. What is the composition of the liquid mixture at 330 K, if the total vapour pressure is 600 mm Hg?
For an ideal binary liquid solutions with P_(A)^(@)gtP_(B)^(@) , which relation between X_(A) (mole fraction of A in liquid phase) and Y_(A) (mole fraction of A in vapour phase) is correct:
The vapour pressure of two pure liquids A and B are 200 and 400 tor respectively at 300K. A liquid solution (ideal) of A and B for which the mole fraction of A is 0.40 is contained in a cylinder. The composition of components A and B in vapour phase after equilibrium is reached between vapour & liquid phase, respectively is
The vapour pressure of pure liquid A and liquid B at 350 K are 440 mm and 720 mm of Hg. If total vapour pressure of solution is 580 mm of Hg then the mole fraction of liquid A in vapour phase will be :-
The vapour pressures of pure liquids A and B are 400 and 600 mm Hg respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are :
The vapour pressures of pure liquids A and B are 400 and 600 mm Hg respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are :
The vapour pressures of pure liquids A and B are 400 and 600 mm Hg respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are :